Supremum

• Jan 5th 2011, 03:00 AM
mus
Supremum
Hi,

Could someone assist me in the right direction concerning the following problem:

Deduce the existence of a supremum from the principle of nested intervals.
• Jan 5th 2011, 04:13 AM
Also sprach Zarathustra
I hope this will help...

Let $(a_n)$ be bounded sequence. Then $S=limsup{a_n}$ exist and he the greater partial limit of the sequence.

In other words: $limsup{a_n}=max\{L:\text{when L is partial limit}\}$

Proof:

Due to Bolzano-Weierstrass Theorem we may discuss on the set of partial limits, which is non-empty.

Given $\epsilon>0$ there are infinite that $S-\epsilon, but only a finite number of $S+\epsilon\leq a_n$. Therefor there is infinite $a_n$ in neighborhood $\epsilon$ of $S$, hence $S$ is partial limit.

Now, suppose that $S other partial limit then we will choose $\epsilon >0$ so that $S, but we know that there is only finite number of $S'-\epsilon, therefor it impossible that $S'$ is partial limit, thus $S$ is the greatest partial limit.
• Jan 5th 2011, 05:12 PM
mus
Could you kindly elaborate on the proof. Its not quite clear.

Thanks